Question: What is the remainder of $8^6 + 7^7 + 6^8$ divided by 5?
First of all, $6$ has a remainder of $1$ when divided by $5,$ thus any power of $6$ will have a remainder of $1$ when divided by $5.$

As for $7,$ it has a remainder of $2$ when divided by $5$, so let us look at the powers of $2$: \begin{align*}
2^1 &\equiv 2 \pmod{5} \\
2^2 &\equiv 4 \pmod{5} \\
2^3 &\equiv 3 \pmod{5} \\
2^4 &\equiv 1 \pmod{5}.
\end{align*}Since $2^4 \equiv 1 \pmod{5},$ we see that $2^{7} \equiv 2^3 \cdot 2^4 \equiv 3 \pmod{5},$ hence the remainder of $7^4$ when divided by $5$ is $3.$

Now, there is a bit of a shortcut we can use for $8.$ Since $8 \equiv -2 \pmod{5},$ we can see that $8^6 \equiv (-2)^6 \equiv 2^6 \equiv 2^2 \cdot 2^4 \equiv 4 \pmod {5},$ hence our desired remainder is $4.$

Adding them up, we have $4 + 3 + 1 \equiv \boxed{3} \pmod{5}.$